The distribution of resistance for resistors of a certain type is known to be normal, with 10% of all resistors having a resistance exceeding 10.256 ohms and 5% having a resistance smaller than 9.671 ohms. What are the mean value and standard deviation of the resistance distribution? slader

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OlaMacgregor OlaMacgregor · Answer 1 · 2020-03-16T11:57:09+01:00

Explanation:

Formula for the given probability is as follows.

[tex]P(z > \frac{x - \mu}{\sigma})[/tex] = 0.10

[tex]P(z > \frac{10.256 - \mu}{\sigma})[/tex] = 0.10

According to the normal table area we have,

P(z < 1.28) = 0.10

Hence, [tex]\frac{10.256 - \mu}{\sigma} = 1.28[/tex]

[tex]\mu = 10.256 - 1.28 \times \sigma[/tex] .......... (1)

Also, the given probability is as follows.

[tex]P(z < \frac{x - \mu}{\sigma})[/tex] = 0.05

[tex]P(z < \frac{9.671 - \mu}{\sigma})[/tex] = 0.05

Hence, [tex]\frac{9.671 - \mu}{\sigma} = -1.645[/tex]

[tex]\mu = 9.671 + 1.645 \times \sigma[/tex] ....... (2)

Now, substitute the value of [tex]\mu[/tex] from equation (1) into equation (2) as follows.

[tex]10.256 - 1.28 \times \sigma = 9.671 + 1.645 \times \sigma[/tex]

[tex]-2.925 \sigma = -0.585[/tex]

[tex]\sigma[/tex] = 0.2

Putting the value of [tex]\sigma[/tex] into equation (2) we will find the value of [tex]\mu[/tex] as follows.

[tex]\mu = 9.671 + 1.645 \times \sigma[/tex]

= [tex]9.671 + 1.645 \times 0.2[/tex]

= 10

Thus, we can conclude that the value of [tex]\sigma[/tex] is 0.2 and the value of [tex]\mu[/tex] is 10.